Question

If $∫\frac{1}{x}$ $√{\frac{1-x}{1+x}}$ _dx = _$g(x) + c,g(1) = 0$ , then g $(\frac{1}{2})$
is equal to

• A. log_e(\frac{√3-1}{√3+1})$$+\frac{π}{3}
• B. log_e(\frac{√3+1}{√3-1})$$+\frac{π}{3}
• C. log_e(\frac{√3+1}{√3-1})$$-\frac{π}{3}
• D. \frac{1}{2}log_e(\frac{√3-1}{√3+1})-$$\frac{π}{6}

Answer 1

Answer: (A)

log_e(\frac{√3-1}{√3+1})$$+\frac{π}{3}

Answer 2

The correct answer is (A) : log_e(\frac{√3-1}{√3+1})$$+\frac{π}{3}

∵ $∫\frac{1}{x}$ $√{\frac{1-x}{1+x}}$ $dx = g(x) + c$
$\int_1^{\frac{1}{2}}\frac{1}{x} \sqrt{\frac{1-x}{1+x}} \,dx = g\left(\frac{1}{2}\right) - g(1)$
∴$g(\frac{1}{2}) =$ $\int_1^{\frac{1}{2}}\frac{1}{x} \sqrt{\frac{1-x}{1+x}} \,dx$
$cotx=cos2θ$
= $\int_0^{\frac{π}{6}} \frac{1}{cos2θ}.\frac{sinθ}{cosθ}(sinθ/cosθ ( -2sin2θ)dθ$
= $-\int_0^{\frac{π}{6}}\frac{4sin²θ}{cos2θ}dθ$
= 2\int_0^{\frac{π}{6}}$$\frac{(1 - 2sin²θ -1)}{cos2θ }$$dθ
= 2 \int^{\frac{π}{6}}_0 $$( 1 - sec2θ ) dθ
=$\frac{π}{3}$ - 2. \frac{1}{2} $$[ In | sec2θ + tan2θ| ]^{\frac{π}{6}}_0
= $\frac{π}{3}$ $- [ In | 2 + √3 | - In1 ]$
= $\frac{π}{3}$ $+ In ( \frac{1}{2 }+ √3 )$
= $\frac{π}{3}$ $+ In | \frac{√3 - 1}{√3 + 1} |$